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Kamila [148]
1 year ago
13

When carbon dioxide dissolves in water, it undergoes a multistep equilibrium process, with Koverall = 4.5x10⁻⁷, which is simplif

ied to the following:(c) What is [CO₃²⁻] in rainwater (Ko of HCO₃⁻ = 4.7 × 10⁻¹¹) ?
Chemistry
1 answer:
seropon [69]1 year ago
3 0

The extensive acid electricity of intact carbonic acid suggests that it is an essential protonation agent below physiological conditions.

<h3>Why is CO2 an acid?</h3>

CO2 is not an acid itself, considering it does no longer incorporate ions of hydrogen (H+). CO2 becomes carbonic acid in water.

Carbonic acid (H2CO3) is a weak, H+-splitting acid. Carbonic acid, a weak acid that acidifies the solution, is formed when some of the carbon dioxide dissolves in the water.

<h3>How do you measure HCO3 in water?</h3>

Since pH = -log [H+], [H+] = 10 to the poor pH. We can use this data and some algebra to rewrite the equation as [HCO3-] = ( two x whole alkalinity ) - ( 10 to the (-14 + pH) ) / ( 1 + 2K2 x 10 to the pH).

Learn more about carbonic acid here:

<h3>brainly.com/question/8552374</h3><h3 /><h3>#SPJ4</h3>

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sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

        CaCO3 + H2SO4 → CaSO4  + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4  × \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}

= 21284.56606

mass of CaCO3 =  2.12 x 10^4 lbs

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid.

To know more about neutralization visit:

brainly.com/question/12498769

#SPJ4

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